Calibration Precision vs Sample Representativeness in HSC Chemistry
Learn why precise AAS measurements do not guarantee a representative sample. Follow worked examples covering blank corrections, calibration limits, dilution factors, and sampling strategies.
An AAS instrument gives almost identical copper concentrations when the same water sample is measured three times. The results are 24.8, 25.0, and 25.2 mg/L. That looks reassuring. But suppose the sample was collected from just one spot beside a drain entering a creek. Can you confidently report that the entire creek contains 25.0 mg/L of copper?
No. The measurements may be precise, but the sample may not be representative.
Calibration precision describes how consistently an analytical method measures known concentrations. Sample representativeness describes how well the collected material reflects the larger material or environment being investigated. These are separate aspects of experimental reliability. A perfectly precise calibration cannot correct a poorly chosen sample.
The distinction becomes much easier once you follow a sample from collection, through preparation, and into the instrument.
01Calibration precision and sample representativeness are different questions
Start by asking what you are actually trying to establish.
Calibration precision: does the measurement repeat consistently?
A calibration curve is constructed using standards containing known concentrations of an analyte.
For example, a chemist might prepare copper standards containing 1.00, 2.00, 4.00, 6.00, and 8.00 mg/L of copper. Each standard is introduced into the instrument, and its absorbance is measured.
If the same 4.00 mg/L standard is measured three times, you might obtain:
| Measurement | Absorbance |
|---|---|
| 1 | 0.198 |
| 2 | 0.200 |
| 3 | 0.202 |
The three readings are close together. Their range is only 0.004 absorbance units.
This suggests that the instrument and calibration procedure produce repeatable responses at this concentration.
A precise calibration should show little variation between repeated measurements of standards at the same concentration. Replicates across the working range provide stronger evidence than repeating just one standard.
However, precision is not the same as accuracy.
A set of standards could be prepared incorrectly but still produce highly consistent readings. The calibration would be precise, but its concentration estimates could be systematically wrong.
A high coefficient of determination, \(R^2\), also does not independently establish accuracy. It describes how closely the calibration data follow the fitted model, not whether the standards were prepared correctly or whether the sample was representative.
Sample representativeness: does the collected material reflect the target?
Now consider the creek.
A drain enters the waterway from the left bank. Water close to the discharge may contain more dissolved copper than water further away.
If you collect one bottle beside the drain, the bottle may accurately represent the water at that particular location. It does not automatically represent the average copper concentration across the creek.
Sample representativeness depends on the sampling strategy.
Consider these two approaches:
- Single-point sampling: Collect one bottle from the bank closest to the discharge. This may be suitable if you specifically want to measure copper concentration at that location.
- Multiple-point sampling: Collect separate bottles from several positions across the creek. This provides information about spatial variation and can help estimate an average concentration.
Neither approach is automatically correct. The appropriate method depends on the question being investigated.
Which experimental result tells you what?
It helps to distinguish four different types of measurements.
| Experimental procedure | What it tests |
|---|---|
| Repeated measurements of the same known standard | Calibration precision |
| Repeated readings from the same prepared sample | Instrument repeatability |
| Independent preparations from the same well-mixed sample | Preparation and measurement precision |
| Independent samples collected from different locations or times | Sample variability and representativeness |
This distinction is important.
Suppose you measure the same sample ten times. You have ten instrument readings, but you still have only one field sample.
You have learned something about measurement repeatability. You have not established how well that bottle represents the entire creek.
02Follow the experiment from collection to measurement
Consider an illustrative investigation designed to determine the copper concentration in a creek.
The monitoring objective is to estimate the mean concentration across three equally weighted positions in the creek at a particular time.
The first experimental design is simple:
- Collect one water sample beside the discharge.
- Bring the sample to the laboratory.
- Dilute a measured portion.
- Measure its absorbance using flame atomic absorption spectroscopy (AAS).
- Repeat the measurement to check consistency.
The prediction is that a stable instrument and calibration should produce similar absorbance readings when the same prepared solution is measured repeatedly.
But there is a problem with this design: it only collects water from one location.
What actually happens inside the instrument?
Understanding the measurement process helps identify exactly what calibration can establish.
In flame AAS, an element-specific lamp produces light at a characteristic wavelength for copper.
A capillary draws the prepared sample solution from a sample container into a nebuliser. The nebuliser produces a fine aerosol, which passes through a spray chamber before entering the flame.
Inside the flame, the sample is atomised. Copper atoms absorb some of the light travelling through the flame.
A detector on the opposite side measures the transmitted light. The instrument uses this measurement to calculate absorbance.
The instrument does not directly measure the copper concentration in the original creek.
It measures the response produced by the prepared solution entering the instrument.
To translate that response into a concentration, the instrument needs a calibration relationship established using known standards.
What does a precise calibration look like?
Suppose the calibration produces the following relationship:
\[
A_{\mathrm{net}}=0.0500c
\]
where:
- \(A_{\mathrm{net}}\) is the blank-corrected absorbance, which is dimensionless.
- \(c\) is the copper concentration in the measured solution, in mg/L.
- \(0.0500\) is the calibration gradient, with units of \(\mathrm{L\,mg^{-1}}\).
The standards have a validated working range of 0.50-8.00 mg/L.
Repeated measurements of the 4.00 mg/L standard produce absorbances of 0.198, 0.200, and 0.202.
These results support the conclusion that the calibration response is repeatable at that concentration.
They do not establish that the original creek sample represents the entire waterway.
That is the experimental limitation.
Revising the sampling design
To address the problem, the monitoring team collects independent samples from three equally weighted positions across the creek.
Each sample is analysed separately.
This revised design allows the team to identify differences between locations and calculate the mean concentration for the defined sampling positions.
If the objective were simply to estimate the mean, an equal-volume composite sample could also be suitable under the appropriate assumptions.
However, if the objective were to identify the highest copper concentration, combining the samples would be a poor choice. The composite would conceal the differences between locations.
The important lesson is that the sampling design must match the quantity being investigated.
03Work from signal to original sample concentration
Once a sample has been collected, the analytical calculation follows a consistent sequence.
\[
\boxed{
\text{Raw signal}
\rightarrow
\text{Blank correction}
\rightarrow
\text{Calibration}
\rightarrow
\text{Dilution correction}
}
\]
Each stage answers a different question.
Step 1: Correct the signal using the blank
A blank contains the solvent and relevant reagents but no target analyte.
It measures background absorbance that should not be attributed to the copper being analysed.
If the sample and blank were measured against the same reference, without first zeroing the instrument using the blank:
\[
A_{\mathrm{net}}
=
A_{\mathrm{sample}}-A_{\mathrm{blank}}
\]
The sign convention matters: subtract the blank from the sample.
For example, if:
\[
A_{\mathrm{sample}}=0.268
\]
and:
\[
A_{\mathrm{blank}}=0.018
\]
then:
\[
A_{\mathrm{net}}=0.268-0.018
\]
\[
A_{\mathrm{net}}=0.250
\]
The corrected absorbance is 0.250.
There is one important qualification. If the instrument has already been zeroed using the blank, the displayed sample absorbance is already corrected. Do not subtract the blank a second time.
Likewise, if the supplied calibration equation uses raw absorbances and already accounts for the background through its intercept, use that equation as supplied rather than applying an additional blank correction.
Step 2: Convert the corrected signal into concentration
For a blank-corrected calibration line passing through the origin:
\[
A_{\mathrm{net}}=mc
\]
Rearrange:
\[
c=\frac{A_{\mathrm{net}}}{m}
\]
The result is the concentration in the solution that entered the instrument.
If the calibration equation has a non-zero intercept \(b\), use:
\[
c=\frac{A_{\mathrm{net}}-b}{m}
\]
Do not assume every calibration curve passes through the origin.
For the full process of reading a calibration graph and finding an unknown concentration, see How to Read an AAS Calibration Curve in HSC Chemistry.
Step 3: Check the measurement limits
A calibration equation can produce a numerical answer even when the measurement is outside the range in which the method has been validated.
That does not make the answer reliable.
Suppose an illustrative analytical method has these stated limits:
| Measurement characteristic | Value |
|---|---|
| Limit of detection (LOD) | 0.10 mg/L |
| Limit of quantification (LOQ) | 0.30 mg/L |
| Lowest calibration standard | 0.50 mg/L |
| Highest calibration standard | 8.00 mg/L |
These values are illustrative method-specific limits, not universal HSC Chemistry limits.
They have different meanings.
- Below 0.10 mg/L: The analyte cannot be reliably distinguished from the background using this method.
- 0.10 to below 0.30 mg/L: The analyte may be detected, but its concentration cannot be reliably quantified.
- 0.30 to below 0.50 mg/L: The result exceeds the stated quantification limit but is below the validated calibration range. The supplied calibration cannot support a quantitative result here.
- 0.50-8.00 mg/L: The result falls within the validated quantitative working range.
- Above 8.00 mg/L: The sample is too concentrated for this calibration. Dilute it further and measure again.
The practical quantitative range is determined by both the quantification limit and the validated calibration range.
Remember that these limits apply to the solution entering the instrument, not necessarily the original sample.
If the original sample was diluted fivefold, the validated working range for the original sample corresponds to:
\[
0.50\times5=2.50\ \mathrm{mg/L}
\]
to:
\[
8.00\times5=40.0\ \mathrm{mg/L}
\]
This assumes no analyte was lost during preparation.
The dilution therefore changes the concentration range that the method can measure in the original sample.
Step 4: Reverse the dilution
Suppose you transfer 10.00 mL of the original sample into a 50.00 mL volumetric flask and add solvent to the calibration mark.
The dilution factor is:
\[
DF=\frac{V_{\mathrm{final}}}{V_{\mathrm{aliquot}}}
\]
where:
- \(V_{\mathrm{final}}\) is the final volume of the diluted solution.
- \(V_{\mathrm{aliquot}}\) is the original sample volume transferred.
Therefore:
\[
DF=\frac{50.00}{10.00}=5.00
\]
The original concentration is:
\[
c_{\mathrm{original}}
=
c_{\mathrm{diluted}}\times DF
\]
If several dilution steps were performed, multiply their individual dilution factors to obtain the total dilution factor.
The final concentration must refer to the original sample if that is what the question asks for.
04Worked examples
Worked example: Calculate the original copper concentration from repeated AAS readings
An analyst collects one bottle of creek water beside a discharge point.
A 10.00 mL aliquot is transferred into a 50.00 mL volumetric flask and diluted to the calibration mark.
The instrument has not been zeroed using the blank. The blank absorbance is 0.018.
Three measurements of the prepared sample give raw absorbances of 0.266, 0.268, and 0.270.
The blank-corrected calibration equation is:
\[
A_{\mathrm{net}}=0.0500c
\]
The validated calibration range is 0.50-8.00 mg/L, with an LOD of 0.10 mg/L and an LOQ of 0.30 mg/L.
Calculate the mean copper concentration in the original creek-water bottle. Explain what the repeated readings establish about the experiment.
Step 1
\[
A_{\mathrm{net}}=A_{\mathrm{sample}}-0.018
\]
This gives:
| Reading | Raw absorbance | Corrected absorbance |
|---|---|---|
| 1 | 0.266 | 0.248 |
| 2 | 0.268 | 0.250 |
| 3 | 0.270 | 0.252 |
Step 2
Rearrange:
\[
c=\frac{A_{\mathrm{net}}}{0.0500}
\]
The three concentrations are:
\[
c_1=\frac{0.248}{0.0500}
=4.96\ \mathrm{mg/L}
\]
\[
c_2=\frac{0.250}{0.0500}
=5.00\ \mathrm{mg/L}
\]
\[
c_3=\frac{0.252}{0.0500}
=5.04\ \mathrm{mg/L}
\]
The mean diluted concentration is:
\[
c_{\mathrm{mean}}=5.00\ \mathrm{mg/L}
\]
All three results are within the validated calibration range.
Step 3
\[
DF=\frac{50.00}{10.00}=5.00
\]
Step 4
\[
c_{\mathrm{original}}
=
5.00\times5.00
\]
\[
\boxed{c_{\mathrm{original}}=25.0\ \mathrm{mg/L}}
\]
The original creek-water bottle contains an estimated 25.0 mg/L of copper.
The three original-sample estimates range from 24.8 to 25.2 mg/L. This small spread indicates good repeatability when measuring the prepared solution.
However, all three readings came from the same field bottle. They do not establish that the bottle represents the mean copper concentration across the creek.
That requires a suitable sampling strategy.
Worked example: Determine whether one sampling location represents the creek
The monitoring team now collects three independent water samples from equally weighted positions across the creek:
- Sample A: near the discharge.
- Sample B: in the middle of the creek.
- Sample C: near the opposite bank.
Each 10.00 mL sample is diluted to 50.00 mL.
The blank absorbance is 0.018, and the calibration equation remains:
\[
A_{\mathrm{net}}=0.0500c
\]
The raw readings are:
| Sampling position | Raw absorbance |
|---|---|
| Near discharge | 0.268 |
| Middle | 0.148 |
| Opposite bank | 0.088 |
Calculate the original copper concentration at each position and determine the mean concentration across the three equally weighted positions.
Step 1
For Sample A:
\[
A_{\mathrm{net,A}}=0.268-0.018=0.250
\]
For Sample B:
\[
A_{\mathrm{net,B}}=0.148-0.018=0.130
\]
For Sample C:
\[
A_{\mathrm{net,C}}=0.088-0.018=0.070
\]
Step 2
Using:
\[
c=\frac{A_{\mathrm{net}}}{0.0500}
\]
we obtain:
| Sample | Corrected absorbance | Diluted concentration |
|---|---|---|
| A | 0.250 | 5.00 mg/L |
| B | 0.130 | 2.60 mg/L |
| C | 0.070 | 1.40 mg/L |
All three diluted concentrations fall within the calibration range of 0.50-8.00 mg/L.
Step 3
For Sample A:
\[
c_A=5.00\times5.00=25.0\ \mathrm{mg/L}
\]
For Sample B:
\[
c_B=2.60\times5.00=13.0\ \mathrm{mg/L}
\]
For Sample C:
\[
c_C=1.40\times5.00=7.00\ \mathrm{mg/L}
\]
Step 4
Because the question specifies three equally weighted positions, use the arithmetic mean:
\[
c_{\mathrm{mean}}
=
\frac{25.0+13.0+7.00}{3}
\]
\[
\boxed{c_{\mathrm{mean}}=15.0\ \mathrm{mg/L}}
\]
The mean copper concentration across the three designated positions is 15.0 mg/L.
Notice what happened.
The original single-point sample gave 25.0 mg/L. That result was not necessarily incorrect. It accurately described the sample collected near the discharge.
The problem would have been reporting that result as the mean concentration across the creek.
The three-location investigation reveals substantial spatial variation.
If the objective were to estimate a flow-weighted average rather than an equally weighted average, the sampling positions would need to be weighted according to the relevant flow contributions. Equal weighting is not automatically appropriate for every waterway.
05A practical decision rule for HSC Chemistry
When a question asks you to assess experimental reliability, identify which part of the experiment is being tested.
| If the question asks… | Look for… |
|---|---|
| Is the calibration precise? | Small variation between repeated measurements of known standards |
| Is the instrument repeatable? | Similar readings from repeated measurements of the same prepared solution |
| Are sample preparations consistent? | Agreement between independently prepared aliquots |
| Is the sample representative? | Appropriate collection locations, depths, times, and sampling strategy |
| Is the concentration quantitatively reliable? | Correct blank treatment, valid calibration range, measurement limits, and dilution corrections |
Four common mistakes are worth avoiding.
Mistake 1: Assuming a precise calibration guarantees a representative sample.
It does not. Calibration precision concerns the measurement process. Representativeness concerns the material collected.
Mistake 2: Treating repeated readings as independent samples.
Measuring one prepared solution five times does not test spatial or temporal variation in the original environment.
Mistake 3: Assuming a high \(R^2\) proves the entire experiment is reliable.
A strong calibration relationship does not establish correct standard preparation, eliminate all systematic errors, or fix poor sampling.
Mistake 4: Using a result outside the calibration range.
A numerical answer from extrapolating beyond the standards is not automatically a valid quantitative result. Dilute an overly concentrated sample and measure it again, or use a suitable method for a sample below the validated range.
06Questions and solutions
Question 1
A student analyses a copper-containing water sample using AAS.
The following information is provided:
- Raw sample absorbance: 0.152
- Blank absorbance: 0.012
- Calibration equation: \(A_{\mathrm{net}}=0.0400c\)
- Sample volume transferred: 5.00 mL
- Final diluted volume: 25.00 mL
- Validated calibration range: 0.50-10.0 mg/L
- LOD: 0.10 mg/L
- LOQ: 0.30 mg/L
The instrument has not been zeroed using the blank.
Calculate the original copper concentration and determine whether the diluted sample falls within the quantitative measurement range.
Solution 1
The original copper concentration is 17.5 mg/L, and the diluted sample is within the validated quantitative range.
Step 1: Correct the absorbance for the blank.
\[
A_{\mathrm{net}}=0.152-0.012
\]
\[
A_{\mathrm{net}}=0.140
\]
Step 2: Calculate the diluted concentration.
Rearrange the calibration equation:
\[
c=\frac{A_{\mathrm{net}}}{0.0400}
\]
Substitute:
\[
c=\frac{0.140}{0.0400}
\]
\[
c=3.50\ \mathrm{mg/L}
\]
Step 3: Calculate the dilution factor.
\[
DF=\frac{25.00}{5.00}=5.00
\]
Step 4: Calculate the original concentration.
\[
c_{\mathrm{original}}=3.50\times5.00
\]
\[
\boxed{c_{\mathrm{original}}=17.5\ \mathrm{mg/L}}
\]
The diluted concentration of 3.50 mg/L is between 0.50 and 10.0 mg/L. It is also above the stated LOQ of 0.30 mg/L.
Therefore, the calibration supports quantitative measurement of this diluted solution.
The final result refers to the original sample, not the diluted solution. Whether that original sample represents a larger water source still depends on how it was collected.
Question 2
A well-mixed field sample is divided into two portions. Each portion is prepared independently using a different dilution.
The calibration equation is:
\[
A_{\mathrm{net}}=0.0250c
\]
The blank absorbance is 0.010. The instrument has not been zeroed using the blank.
The results are:
| Measurement | Preparation A | Preparation B |
|---|---|---|
| Original aliquot | 10.00 mL | 5.00 mL |
| Final volume | 50.00 mL | 40.00 mL |
| Raw absorbance | 0.210 | 0.135 |
The validated calibration range is 0.50-10.0 mg/L.
Calculate the original copper concentration using each preparation. Explain why the raw absorbance readings are different and what the agreement between the final concentrations tells you about the experiment.
Solution 2
Both preparations give an original copper concentration of 40.0 mg/L. The different absorbances are explained by the different dilution factors.
Step 1: Correct both absorbances for the blank.
Preparation A:
\[
A_{\mathrm{net,A}}=0.210-0.010
\]
\[
A_{\mathrm{net,A}}=0.200
\]
Preparation B:
\[
A_{\mathrm{net,B}}=0.135-0.010
\]
\[
A_{\mathrm{net,B}}=0.125
\]
Step 2: Calculate the diluted concentrations.
For Preparation A:
\[
c_A=\frac{0.200}{0.0250}
\]
\[
c_A=8.00\ \mathrm{mg/L}
\]
For Preparation B:
\[
c_B=\frac{0.125}{0.0250}
\]
\[
c_B=5.00\ \mathrm{mg/L}
\]
Both concentrations are within the validated calibration range.
Step 3: Calculate the individual dilution factors.
Preparation A:
\[
DF_A=\frac{50.00}{10.00}=5.00
\]
Preparation B:
\[
DF_B=\frac{40.00}{5.00}=8.00
\]
Step 4: Recover the original concentration.
Preparation A:
\[
c_{\mathrm{original,A}}=8.00\times5.00
\]
\[
c_{\mathrm{original,A}}=40.0\ \mathrm{mg/L}
\]
Preparation B:
\[
c_{\mathrm{original,B}}=5.00\times8.00
\]
\[
c_{\mathrm{original,B}}=40.0\ \mathrm{mg/L}
\]
Therefore:
\[
\boxed{c_{\mathrm{original}}=40.0\ \mathrm{mg/L}}
\]
The raw absorbances differ because the solutions entering the instrument have different concentrations.
Preparation A is diluted fivefold, while Preparation B is diluted eightfold. The more diluted solution produces the smaller absorbance.
The agreement between the original concentrations supports consistency between the independent preparations and measurements.
However, both portions came from the same field bottle. Their agreement does not establish that the original bottle represents the entire sampling location.
Question 3
A chemist analyses a water sample using the following method:
- First dilution: 2.00 mL of original sample made up to 20.00 mL.
- Initial raw absorbance: 0.438.
- Blank absorbance: 0.018.
- Calibration equation: \(A_{\mathrm{net}}=0.0500c\).
- Validated calibration range: 0.50-8.00 mg/L.
- LOQ: 0.30 mg/L.
The initial absorbance is measured outside the validated calibration range.
The chemist then takes 2.00 mL of the first diluted solution and makes it up to 10.00 mL. The second measurement produces a raw absorbance of 0.102.
Calculate the original sample concentration using the second measurement. Explain why the initial calculated concentration should not be reported.
Solution 3
The original copper concentration is 84.0 mg/L, calculated using the second measurement, which falls within the validated calibration range.
Step 1: Examine the initial measurement.
The first dilution factor is:
\[
DF_1=\frac{20.00}{2.00}=10.0
\]
The corrected initial absorbance is:
\[
A_{\mathrm{net}}=0.438-0.018
\]
\[
A_{\mathrm{net}}=0.420
\]
Using the calibration equation:
\[
c=\frac{0.420}{0.0500}
\]
\[
c=8.40\ \mathrm{mg/L}
\]
This result exceeds the upper calibration limit of 8.00 mg/L.
Although the equation produces a numerical value, it cannot support a reliable quantitative result outside the validated range.
The chemist must dilute the sample further.
Step 2: Calculate the second dilution factor.
The second dilution transfers 2.00 mL into a final volume of 10.00 mL.
\[
DF_2=\frac{10.00}{2.00}=5.00
\]
Step 3: Correct the second absorbance.
\[
A_{\mathrm{net}}=0.102-0.018
\]
\[
A_{\mathrm{net}}=0.084
\]
Step 4: Calculate the concentration in the second diluted solution.
\[
c=\frac{0.084}{0.0500}
\]
\[
c=1.68\ \mathrm{mg/L}
\]
This result is within the validated calibration range and above the LOQ.
Step 5: Calculate the total dilution factor.
The dilution factors are multiplied because the second dilution was performed on the already diluted solution.
\[
DF_{\mathrm{total}}=DF_1\times DF_2
\]
\[
DF_{\mathrm{total}}=10.0\times5.00
\]
\[
DF_{\mathrm{total}}=50.0
\]
Step 6: Recover the original concentration.
\[
c_{\mathrm{original}}=1.68\times50.0
\]
\[
\boxed{c_{\mathrm{original}}=84.0\ \mathrm{mg/L}}
\]
The second measurement provides the defensible quantitative result because it falls within the calibrated range.
The initial result of 8.40 mg/L was useful as an indication that further dilution was needed. It was not a valid final measurement.
Question 4
An industrial process produces wastewater whose copper concentration changes throughout an eight-hour shift.
The monitoring objective is to estimate the mean concentration across four equal two-hour periods. For this question, a sample collected at the midpoint of each period is assumed to represent that period.
An analyst collects a sample at 9 am and measures it five times. After blank correction, calibration, and dilution correction, the results are:
\[
9.9,\ 10.0,\ 10.0,\ 10.1,\ 10.0\ \mathrm{mg/L}
\]
A second analyst collects independent samples at the midpoint of each two-hour period. The resulting original-sample concentrations are:
| Sampling time | Copper concentration |
|---|---|
| 9 am | 10.0 mg/L |
| 11 am | 14.0 mg/L |
| 1 pm | 22.0 mg/L |
| 3 pm | 14.0 mg/L |
Which analyst provides the more appropriate estimate of the defined eight-hour mean concentration?
Calculate the mean and explain why the first analyst’s repeated readings are insufficient, even though they show good precision.
Solution 4
The second analyst provides the appropriate estimate: 15.0 mg/L. The first analyst has demonstrated measurement repeatability at one time, not representativeness across the entire shift.
Step 1: Calculate the mean from the four independent sampling times.
The four periods have equal durations and are equally weighted.
Therefore:
\[
c_{\mathrm{mean}}
=
\frac{10.0+14.0+22.0+14.0}{4}
\]
\[
\boxed{c_{\mathrm{mean}}=15.0\ \mathrm{mg/L}}
\]
Step 2: Interpret the first analyst’s results.
The five readings from the 9 am sample have a mean of:
\[
\frac{9.9+10.0+10.0+10.1+10.0}{5}
=
10.0\ \mathrm{mg/L}
\]
The narrow spread indicates good measurement repeatability.
However, all five readings came from the same sample collected at 9 am.
They provide no information about the changes occurring later in the shift.
Step 3: Explain the sampling limitation.
The concentration increases to 22.0 mg/L at 1 pm before decreasing again.
A single morning sample cannot capture this variation.
The second analyst’s sampling strategy better matches the defined time-averaged quantity because it includes independent samples across all four periods.
The important distinction is that repeated measurements improve evidence about precision, while collecting samples at appropriate times improves evidence about temporal representativeness.
Question 5
A monitoring team is investigating copper concentration in a storage tank divided into three equally weighted sampling zones.
The objective is to determine whether any one of the three zones exceeds an illustrative internal action trigger of 20.0 mg/L. This trigger is supplied for the scenario and is not a stated regulatory limit.
A technician collects 10.00 mL from each zone and combines the three portions into one composite sample.
The technician then transfers 5.00 mL of the composite into a 25.00 mL volumetric flask and dilutes it to the calibration mark.
The AAS measurement gives:
- Raw sample absorbance: 0.140
- Blank absorbance: 0.012
- Calibration equation: \(A_{\mathrm{net}}=0.0400c\)
- Validated calibration range: 0.50-8.00 mg/L
- LOD: 0.10 mg/L
- LOQ: 0.30 mg/L
The technician calculates the composite concentration and reports:
“The copper concentration is below 20.0 mg/L, so all three zones are below the action trigger.”
Evaluate this conclusion.
Calculate the composite concentration, explain what the measurement actually establishes, and recommend a sampling strategy that would answer the original monitoring question.
Solution 5
The composite concentration is 16.0 mg/L, but the technician’s conclusion is invalid. A composite measurement estimates the equally weighted mean concentration, not the maximum concentration in any individual zone.
Step 1: Correct the absorbance for the blank.
\[
A_{\mathrm{net}}=0.140-0.012
\]
\[
A_{\mathrm{net}}=0.128
\]
Step 2: Calculate the diluted composite concentration.
Rearrange the calibration equation:
\[
c=\frac{A_{\mathrm{net}}}{0.0400}
\]
Substitute:
\[
c=\frac{0.128}{0.0400}
\]
\[
c=3.20\ \mathrm{mg/L}
\]
The diluted concentration is within the validated range of 0.50-8.00 mg/L and above the LOQ of 0.30 mg/L.
Therefore, the instrument measurement supports quantitative analysis of the composite.
Step 3: Calculate the dilution factor.
\[
DF=\frac{25.00}{5.00}=5.00
\]
Step 4: Recover the composite concentration.
\[
c_{\mathrm{composite}}=3.20\times5.00
\]
\[
\boxed{c_{\mathrm{composite}}=16.0\ \mathrm{mg/L}}
\]
The equal-volume composite represents the arithmetic mean of the three equally weighted zones, assuming no analyte is lost during collection, mixing, or preparation.
But the mean does not reveal the individual concentrations.
Consider two possible sets of concentrations:
| Possible distribution | Zone 1 | Zone 2 | Zone 3 | Mean |
|---|---|---|---|---|
| Distribution A | 16.0 | 16.0 | 16.0 | 16.0 |
| Distribution B | 30.0 | 12.0 | 6.0 | 16.0 |
All concentrations are in mg/L.
Both distributions produce exactly the same composite mean.
However, Distribution A has no zone above 20.0 mg/L, while Distribution B has one zone at 30.0 mg/L.
The composite result cannot distinguish between these possibilities.
Step 5: Choose a sampling method that matches the objective.
The original question asks whether any individual zone exceeds the action trigger.
The appropriate strategy is therefore to:
- Collect separate samples from all three zones.
- Keep the samples separate during preparation.
- Apply the appropriate blank correction to each measurement.
- Use the calibration equation to calculate each diluted concentration.
- Check that each measurement is within the validated range.
- Apply the appropriate dilution factor to recover each original concentration.
- Compare each individual result with 20.0 mg/L.
If the technician has already combined the samples and discarded the individual portions, the spatial information has been lost. The three zones must be sampled again to answer the original question.
There is an important distinction here.
The composite sample is potentially representative of the mean concentration. It is not representative of the highest concentration or the individual concentrations within the tank.
A sample can be suitable for one analytical objective and unsuitable for another, even when its measurement is perfectly precise.
07The next decision: what are you trying to represent?
When interpreting an analytical result, separate three decisions.
First, determine whether the calibration provides a repeatable relationship between signal and known concentration.
Second, determine whether the collected material represents the specific location, batch, time period, or average you want to investigate.
Third, establish whether the measured signal supports a quantitative result after blank correction, calibration, measurement-limit checks, and dilution correction.
These decisions should not be treated as interchangeable.
If you are comparing two independently prepared sample concentrations, the next useful skill is understanding how to interpret their differences. See How to Compare Duplicate Sample Concentrations in HSC Chemistry.
A precise measurement tells you how consistently the instrument can measure a prepared sample. A representative sampling strategy tells you whether that sample can answer the question you originally asked.